A transverse harmonic wave on a string is described by, \(y(x,t) = 3.0 ~\sin \left(36t + 0.018x + {\dfrac {\pi} 4}\right)\) where \(x\) and \(y\)are in cm and \(t\)in sec. The positive direction of \(x\) is from left to right. What is the shortest distance between two successive crests in the wave?
1.
\(1.3\) m
2.
\(3.0\) m
3.
\(2.5\) m
4.
\(3.5\) m
Add Note
Subtopic: Â Wave Motion |
 68%
From NCERT
Other Reason
Highlight in NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Upgrade Your Plan
Upgrade now and unlock your full question bank experience with daily practice.
For the travelling harmonic wave, \(y(x,t) = 2.0\cos 2\pi (10t - 0.0080x + 0.35 )\) where \(x\) and \(y\) are in \(\text{cm}\) and \(t\) is in seconds. The phase difference between the oscillatory motion of two points separated by a distance of \(4~\text{m}\) will be:
1. \(0.8 \pi~\text{rad}\)
2.\(\pi~ \text{rad}\)
3. \(6.4\pi~\text{rad}\)
4. \(4\pi~\text{rad}\)
Add Note
Subtopic: Â Wave Motion |
 68%
From NCERT
Other Reason
Highlight in NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Upgrade Your Plan
Upgrade now and unlock your full question bank experience with daily practice.